Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define pb push_back
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define mem(a,k) memset(a,k,sizeof(a))
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#define Pr(f,...) //fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)

typedef long long ll;

template<typename T>
void read(T &x){
	x=0; 
	char c;
	for(c=getchar();!isdigit(c);c=getchar()); 
	for(;isdigit(c);c=getchar())x=x*10+c-'0'; 
}

const int N=2e5+50;

int n,p[N],f[N],T[N],q[N]; 

bool chk(int a,int b,int c){
	return (a<b&&b<c)||(b<c&&c<a)||(c<a&&a<b); 
}

int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
	//freopen("dat.in","r",stdin);
	//freopen("my.out","w",stdout);
#endif
	read(n);
	rep(i,1,n){
		read(p[i]);
		q[p[i]]=i;
	}
	T[n]=0,f[n]=n;
	dec(i,n-1,1){
		if(T[i+1]==0){
			if(q[i]>q[i+1]){
				T[i]=1;
				f[i]=i+1; 
			}else{
				T[i]=0; 
			}
		}else{
			if(chk(q[f[i+1]],q[i],q[i+1])){
				T[i]=T[i+1]; 
				f[i]=f[i+1];
			}else{
				T[i]=T[i+1]+1;
				f[i]=i+1; 
			}
		}
	}
	printf("%d\n",T[1]); 
	return 0;
}